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  "body": "\n## Statement\n\nLet $f \\colon (0, \\infty) \\to \\C$ be measurable, let $a \\in \\C$, and let $g(x) = x^{a} f(x)$ for $x \u003e 0$, where $x^{a} = e^{a \\log x}$ is the [[complex power|0RNGDMHT7M]].\nLet $s \\in \\C$.\nThen $g$ is [[Mellin convergent|SGZZ4Y0D36]] at $s$ if and only if $f$ is Mellin convergent at $s + a$.\nIn that case\n$$\\mathcal{M}g(s) = \\mathcal{M}f(s + a),$$\nwhere $\\mathcal{M}$ is the [[Mellin transform|N5PVGRJE1T]].\n\n## Remarks\n\nThe proof is the identity $x^{s-1} \\cdot x^{a} = x^{(s + a) - 1}$ for $x \u003e 0$, which holds because $\\log x$ is real.\nTaking absolute values gives $\\abs{x^{s-1} g(x)} = x^{\\operatorname{Re}(s + a) - 1} \\abs{f(x)}$, hence the equivalence of convergence.\nSo the strip of convergence is shifted by $-\\operatorname{Re} a$.\nExample: $f(x) = e^{-x}$ and $a = 1$ give $\\mathcal{M}[x e^{-x}](s) = \\Gamma(s + 1)$ for $\\operatorname{Re} s \u003e -1$.\n\n## Explanations\n\n### L1 human\n\nMultiplying a function by a power $x^{a}$ just shifts its Mellin transform by $a$ in the variable $s$.\nThis is the Mellin counterpart of the rule that multiplying by $e^{i\\omega x}$ shifts a Fourier transform.\n\n### L2 human\n\nThe rule shows that the choice of the exponent $s - 1$ rather than $s$ in the Mellin kernel is a convention: changing it only shifts $s$.\nWith the dilation rule and the power substitution rule, it gives the Mellin transforms of all functions $x^{b} f(c x^{a})$ from that of $f$.\nIt is also how a weight $x^{\\sigma}$ is moved in and out of Plancherel-type identities for the Mellin transform.\n",
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